WAEC SSCE - General Mathematics - 2005
Tambaya 1 Rahoto
The lengths of the adjacent sides of a right - angled triangle are xcm, (x-1)cm. If the length of the hypotenuse is \(\sqrt{13}cm\), find the value of x
Bayanin Amsa
Let's use the Pythagorean Theorem which states that in a right-angled triangle, the square of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides. Therefore, for this triangle, we have: \begin{align*} (\text{hypotenuse})^2 &= (\text{one side})^2 + (\text{other side})^2 \\ (\sqrt{13})^2 &= x^2 + (x-1)^2 \\ 13 &= x^2 + (x-1)^2 \\ 13 &= x^2 + x^2 - 2x + 1 \\ 0 &= 2x^2 - 2x - 12 \\ 0 &= x^2 - x - 6 \\ 0 &= (x-3)(x+2) \end{align*} We get two values: x=3 and x=-2. However, x must be positive, so we choose x=3 as the correct value. Therefore, the value of x is 3.
Tambaya 2 Rahoto
In the diagram, PQRS is a parallelogram and ?QRT = 30o. Find x
Tambaya 3 Rahoto
Correct 0.04945 to two significant figures
Bayanin Amsa
When rounding a number to a certain number of significant figures, we look at the digit to the right of the last significant figure we want to keep. If that digit is 5 or greater, we round up the last significant figure. If it is less than 5, we simply drop all the digits to the right of the last significant figure. In this case, we want to round 0.04945 to two significant figures. The last significant figure we want to keep is the 9, so we look at the digit to its right, which is 4. Since 4 is less than 5, we simply drop all the digits to the right of the last significant figure, and the rounded value is 0.049. Therefore, the correct answer is option (ii) 0.049.
Tambaya 4 Rahoto
A train moving at a uniform speed covers 36km in 21 minutes. How long does it take to cover 60km?
Bayanin Amsa
We can use the formula:
distance = speed x time
to solve the problem.
We are given that the train covers 36 km in 21 minutes. We want to find how long it will take to cover 60 km.
Let's first convert 21 minutes to hours:
21 minutes = 21/60 hours = 0.35 hours
Now we can use the formula to find the speed of the train:
36 km = speed x 0.35 hours
Solving for speed, we get:
speed = 36/0.35 km/h
Using a calculator, we can evaluate this expression to get:
speed = 102.86 km/h
Now we can use the speed to find the time it takes to cover 60 km:
60 km = 102.86 km/h x time
Solving for time, we get:
time = 60/102.86 hours
Converting this to minutes, we get:
time = (60/102.86) x 60 minutes
Simplifying this expression, we get:
time = 35 minutes (approx)
Therefore, the train will take approximately 35 minutes to cover 60 km.
So the correct option is 1.
Tambaya 5 Rahoto
The figure shows a quadrilateral PQRS having equal sides and opposite sides parallel. The diagonals PR and QS intersect perpendicularly at O, Which of the following statements cannot be correct
Bayanin Amsa
Since PQRS is a quadrilateral with equal sides and opposite sides parallel, it must be a parallelogram. Since PR and QS are diagonals of this parallelogram and intersect perpendicularly at O, O is the midpoint of both PR and QS. Therefore, |PO|=|RO|. If PQR is an equilateral triangle, then PQ=QR=RP, which means that PQRS would be a rhombus (a special case of a parallelogram with all sides equal), not just a parallelogram. This contradicts the given information, so option (B) cannot be correct. If PQRX is a parallelogram, then PQ and RX are parallel and PQ=RX because PQRS is a parallelogram with opposite sides parallel and equal in length. This would mean that the diagonals PR and QS intersect at the midpoint of PQ and RX, not perpendicularly. This contradicts the given information that PR and QS intersect perpendicularly at O, so option (C) cannot be correct. There are no conditions given about the symmetry of the quadrilateral PQRS, so option (D) may or may not be correct. The presence or absence of lines of symmetry does not contradict any of the given information or the other options. Therefore, the statement that cannot be correct is option (B), "PQR is an equilateral triangle," since it contradicts the given information that PQRS is a parallelogram with equal sides and opposite sides parallel.
Tambaya 7 Rahoto
A regular polygon has 9 sides. What is the size of one of its exterior angles?
Bayanin Amsa
The sum of the exterior angles of any polygon is 360 degrees. In a regular polygon, all the exterior angles have the same measure, so to find the measure of one exterior angle, we divide 360 by the number of sides. For a regular polygon with 9 sides, each exterior angle has a measure of 360/9 = 40 degrees. Therefore, the answer is option (B), 40o. To summarize, the measure of one exterior angle of a regular polygon with 9 sides is 40 degrees, since the sum of the exterior angles is 360 degrees and they are all equal in measure.
Tambaya 8 Rahoto
Solve the equation 2x - 3y = 22; 3x + 2y = 7
Bayanin Amsa
To solve the system of equations 2x - 3y = 22 and 3x + 2y = 7, we can use the method of elimination. Multiplying the first equation by 2 and the second equation by 3, we get: 4x - 6y = 44 (1) 9x + 6y = 21 (2) Adding equations (1) and (2), we get: 13x = 65 Dividing both sides by 13, we get: x = 5 Substituting x = 5 into the second equation, we get: 3(5) + 2y = 7 Simplifying: 15 + 2y = 7 Subtracting 15 from both sides, we get: 2y = -8 Dividing both sides by 2, we get: y = -4 Therefore, the solution to the system of equations is x = 5 and y = -4, which corresponds to option (D).
Tambaya 10 Rahoto
A sequence is given by \(2\frac{1}{2}, 5, 7\frac{1}{2}, .....\) if the nth term is 25, find n
Bayanin Amsa
The given sequence is an arithmetic sequence, because the difference between consecutive terms is constant. To find the common difference d, we subtract the second term from the first term:
5 - 2.5 = 2.5
So the common difference is 2.5. We can use this common difference to find any term of the sequence, using the formula:
an = a1 + (n - 1)d
where an is the nth term, a1 is the first term, n is the term number, and d is the common difference.
We know that the nth term is 25, so we can set an = 25 and solve for n:
25 = 2.5 + (n - 1)2.5
Simplifying the equation, we get:
22.5 = 2.5n - 2.5 25 = 2.5n n = 10
Therefore, the term number n for which the nth term of the sequence is 25 is 10, which corresponds to.
Therefore, the correct answer is: 10.
Tambaya 11 Rahoto
A fair die is tossed once, what is the probability of obtaining neither 5 or 2
Bayanin Amsa
A fair die has six equally likely outcomes, which are the numbers 1, 2, 3, 4, 5, and 6. To find the probability of obtaining neither 5 nor 2, we first need to find the number of outcomes that satisfy this condition. There are four outcomes that are not 5 or 2, which are 1, 3, 4, and 6. Therefore, the probability of obtaining neither 5 nor 2 is the number of outcomes that satisfy this condition divided by the total number of possible outcomes, which is 4/6 or 2/3. Thus, the answer is option (B), \(\frac{2}{3}\). To summarize, the probability of obtaining neither 5 nor 2 when a fair die is tossed once is 2/3, since there are four outcomes that satisfy this condition out of a total of six possible outcomes.
Tambaya 12 Rahoto
Evaluate, correct to the nearest whole number \(7\frac{1}{2}-\left(2\frac{1}{2}+3\right)\div\frac{33}{2}\)
Bayanin Amsa
To evaluate the expression, we must follow the order of operations, which is also known as PEMDAS (parentheses, exponents, multiplication and division, addition and subtraction). Starting with the parentheses, we have: $$7\frac{1}{2}-\left(2\frac{1}{2}+3\right)\div\frac{33}{2}$$ $$=7\frac{1}{2}-\left(5+\frac{3}{1}\right)\div\frac{33}{2}$$ $$=7\frac{1}{2}-\left(5+\frac{3}{1}\right)\times\frac{2}{33}$$ $$=7\frac{1}{2}-\left(5\times\frac{2}{33}+\frac{3}{1}\times\frac{2}{33}\right)$$ $$=7\frac{1}{2}-\left(\frac{10}{33}+\frac{6}{33}\right)$$ $$=7\frac{1}{2}-\frac{16}{33}$$ Now, we need to find a common denominator to subtract the fractions. We can convert the mixed number to an improper fraction and multiply by $\frac{33}{33}$ to get: $$=7\frac{1}{2}\times\frac{33}{33}-\frac{16}{33}$$ $$=\frac{15}{2}\times\frac{33}{33}-\frac{16}{33}$$ $$=\frac{495}{66}-\frac{16}{33}$$ $$=\frac{495-32}{66}$$ $$=\frac{463}{66}$$ To round to the nearest whole number, we divide 463 by 66 and round to the nearest whole number: $$\frac{463}{66}\approx7$$ Therefore, the correct answer is 7.
Tambaya 13 Rahoto
XOY is a real sector of a circle center O of radius 3.5cm which subtends an angle of 144o at the center. Calculate, in term of ϖ, the area of the sector
Bayanin Amsa
To find the area of the sector, we first need to find the length of the arc XOY.
The formula for the length of an arc of a sector is given by:
length of arc = (angle/360) x 2 x ϖ x radius
Substituting the given values, we get:
length of arc = (144/360) x 2 x ϖ x 3.5
Simplifying this expression, we get:
length of arc = 1.4ϖ
So the length of arc XOY is 1.4ϖ cm.
To find the area of the sector, we use the formula:
area of sector = (angle/360) x ϖ x radius^2
Substituting the given values, we get:
area of sector = (144/360) x ϖ x 3.5^2
Simplifying this expression, we get:
area of sector = 4.9ϖ
Therefore, the area of the sector is 4.9ϖ cm2.
So the correct option is 4.
Tambaya 14 Rahoto
In the diagram PQRS is a circle, |PT| = |QT| and ?QPT = 70o what is the size of ?PRS?
Bayanin Amsa
Tambaya 15 Rahoto
What is the equation of the curve?
Bayanin Amsa
The equation of the curve is: \(y=x^2-5x+4\) This is a quadratic equation in standard form, where the coefficient of the \(x^2\) term is 1, the coefficient of the \(x\) term is -5, and the constant term is 4. The graph of a quadratic equation is a parabola. The coefficient of the \(x^2\) term determines whether the parabola opens upwards or downwards. In this case, since the coefficient is positive, the parabola opens upwards. To find the vertex of the parabola, we can use the formula: \(-\frac{b}{2a}\), where \(a\) is the coefficient of the \(x^2\) term and \(b\) is the coefficient of the \(x\) term. Substituting the values, we get: \(-\frac{b}{2a}=-\frac{-5}{2(1)}=\frac{5}{2}\) So the vertex is at \(\left(\frac{5}{2},\frac{1}{4}\right)\). Therefore, the equation of the curve is \(y=x^2-5x+4\).
Tambaya 16 Rahoto
The salary of a man was increased in the ratio 40:47. calculate the percentage increase in the salary
Bayanin Amsa
To calculate the percentage increase in the salary of a man, we need to find the difference between the new and old salary and then express it as a percentage of the old salary. Let's assume the old salary was 40x and the new salary is 47x (where x is a constant). Then, the increase in salary = 47x - 40x = 7x. Now, the percentage increase in salary = (increase in salary / old salary) x 100% = (7x / 40x) x 100% = 17.5% Therefore, the percentage increase in the salary of the man is 17.5%. Answer is correct.
Tambaya 17 Rahoto
A rectangular packet has inner dimension 16cm by 12cm by 6cm. How many cubes of sugar of side 2cm can be neatly packed into the packet?
Bayanin Amsa
To solve this problem, we need to calculate the volume of the rectangular packet and the volume of each cube of sugar, and then divide the volume of the packet by the volume of each cube. The volume of the rectangular packet is: $$V_{packet} = 16 \text{cm} \times 12 \text{cm} \times 6 \text{cm} = 1152 \text{cm}^3$$ The volume of each cube of sugar is: $$V_{cube} = 2 \text{cm} \times 2 \text{cm} \times 2 \text{cm} = 8 \text{cm}^3$$ To calculate how many cubes of sugar can be packed into the packet, we divide the volume of the packet by the volume of each cube: $$\frac{V_{packet}}{V_{cube}} = \frac{1152 \text{cm}^3}{8 \text{cm}^3} = 144$$ Therefore, 144 cubes of sugar of side 2cm can be neatly packed into the rectangular packet. Hence, the correct answer is 144.
Tambaya 18 Rahoto
Convert 101101two to a number in base ten
Bayanin Amsa
To convert a binary number to a decimal number, we need to multiply each digit of the binary number by its corresponding power of 2, starting from the rightmost digit with the power of 0, and then add the products.
Let's apply this method to convert the binary number 101101 to a decimal number:
1 0 1 1 0 1 (binary digits) 25 24 23 22 21 20 (corresponding powers of 2) 32 0 8 4 0 1 (products)
Now we can add the products to get the decimal equivalent of the binary number:
32 + 0 + 8 + 4 + 0 + 1 = 45
Therefore, the binary number 101101two is equal to the decimal number 45, which corresponds to.
Therefore, the correct answer is: 45.
Tambaya 19 Rahoto
In the diagram, O is the center of the circle, ?MON = 80o, ?LMO = 10o and ?LNO = 15o. Calculate the value of x
Bayanin Amsa
Tambaya 20 Rahoto
The probability that John and James pass an examination are 3/4 and 3/5 respectively, find the probability of both boys failing the examination
Bayanin Amsa
The probability that John passes the exam is 3/4, so the probability that John fails the exam is 1-3/4=1/4. Similarly, the probability that James fails the exam is 1-3/5=2/5. Now, we need to find the probability that both John and James fail the exam. We can use the multiplication rule of probability which states that the probability of two independent events occurring together is the product of their individual probabilities. Therefore, the probability of both John and James failing the exam is: (1/4) x (2/5) = 1/10 So, the answer is \(\frac{1}{10}\).
Tambaya 21 Rahoto
Find the mean of the numbers 1, 3, 4, 8, 8, 4 and 7
Bayanin Amsa
To find the mean of a set of numbers, we add up all the numbers and then divide the sum by the total number of numbers. So, in this case, we add up the numbers: $$1+3+4+8+8+4+7=35$$ There are 7 numbers in the set, so we divide the sum by 7 to get: $$\frac{1+3+4+8+8+4+7}{7}=\frac{35}{7}=5$$ Therefore, the mean (also known as the average) of the numbers 1, 3, 4, 8, 8, 4, and 7 is 5. So, the correct answer is 5.
Tambaya 22 Rahoto
What is the total surface area of a closed cylinder of height 10cm and diameter 7cm? [Take \(\pi = \frac{22}{7}\)]
Bayanin Amsa
The formula for the total surface area of a cylinder is given by: `2πrh + 2πr^2` Where `r` is the radius of the circular base of the cylinder and `h` is its height. In this case, the diameter of the cylinder is given as 7cm. We need to find the radius of the cylinder first. We know that the diameter is 7cm, which means the radius is half of the diameter. `radius = 7/2 = 3.5cm` The height of the cylinder is given as 10cm. Now, we can substitute the values in the formula to find the total surface area of the cylinder. `2πrh + 2πr^2` `= 2 × (22/7) × 3.5 × 10 + 2 × (22/7) × 3.5^2` `= 220 + 77` `= 297cm^2` Therefore, the total surface area of the cylinder is 297cm^2. The answer is.
Tambaya 23 Rahoto
The monthly salary of a man increased from N2,700 to N3,200. Find the percentage increase
Bayanin Amsa
We can use the formula for percentage increase: percentage increase = (new value - old value) / old value x 100% Here, the old salary was N2,700 and the new salary is N3,200. So, the percentage increase in salary would be: = (3200 - 2700) / 2700 x 100% = 500 / 2700 x 100% = 0.185 x 100% = 18.5% Therefore, the percentage increase in the man's salary is 18.5%. Hence, the answer is: 18.5%.
Tambaya 24 Rahoto
If \(log_9x= 1.5\),find x
Bayanin Amsa
We can rewrite the equation in exponential form as follows: $$9^{1.5} = x$$ Using the exponent rules, we can simplify $9^{1.5}$ as: $$9^{1.5} = \sqrt{9^3} = 3^3 = 27$$ Therefore, $x = 27$. So the correct option is (B) 27.
Tambaya 26 Rahoto
P = {3, 9, 11, 13} and Q = {3, 7, 9, 15} are subset of the universal set ξ = {1, 3, 7, 9, 11, 13, 15} find PI ∩ QI
Bayanin Amsa
First, let's find PI, which is the complement of P in ξ: PI = ξ \ P = {1, 7, 15} Similarly, QI = ξ \ Q = {1, 11, 13} The intersection of PI and QI is the set of elements that are in both sets. In this case, the intersection is: PI ∩ QI = {1} Therefore, the answer is option (C) {1}.
Tambaya 27 Rahoto
From a point R, 300m north of P, man walks eastward to a place Q which is 600m from P. Find the bearing of P from Q, correct to the nearest degreeee
Bayanin Amsa
Tambaya 29 Rahoto
expand (2x-3y)(x-5y)
Bayanin Amsa
To expand (2x-3y)(x-5y), we use the distributive property of multiplication, which states that a(b+c) = ab + ac. Applying this property, we get: (2x-3y)(x-5y) = 2x(x-5y) - 3y(x-5y) = 2x^2 - 10xy - 3xy + 15y^2 = 2x^2 - 13xy + 15y^2 Therefore, the answer is (2x^2 - 13xy + 15y^2).
Tambaya 30 Rahoto
Ladi sold a car for N84,000 at a loss of 4%. How much did ladi buy the car
Bayanin Amsa
When Ladi sold the car, he made a loss of 4%. This means that he sold the car for 96% of its original value (100% - 4% = 96%). We can use this information to find the original price of the car. Let's assume that the original price of the car was x Naira. Then, the selling price of the car (i.e., the price at which Ladi sold the car) would be 96% of x, or 0.96x Naira. We know that Ladi sold the car for N84,000, so we can set up an equation: 0.96x = 84,000 Solving for x, we get: x = 84,000 / 0.96 x = 87,500 Therefore, the original price of the car was N87,500. So the correct option is 4.
Tambaya 32 Rahoto
For what values of y is the expression \(\frac{6y-1}{y^2 - y-6}\)
Bayanin Amsa
To determine the values of y for which the expression \(\frac{6y-1}{y^2 - y-6}\) is defined, we need to find the values of y that make the denominator of the expression nonzero. The denominator of the expression is \(y^2 - y-6\), which can be factored as \((y-3)(y+2)\). Therefore, the expression is undefined when \(y=3\) or \(y=-2\), because in those cases the denominator becomes zero. So, the values of y for which the expression is defined are all the real numbers except 3 and -2. That means, y can be any number other than 3 and -2. Therefore, the correct answer is: y can be any number other than 3 and -2.
Tambaya 33 Rahoto
Find the equation whose roots are 2 and \(-3\frac{1}{2}\)
Tambaya 34 Rahoto
What is the diameter of a circle of area 77cm2 [Take \(\pi = \frac{22}{7}\)]
Bayanin Amsa
Tambaya 35 Rahoto
What are the roots of the equation?
Tambaya 37 Rahoto
A man bought 220 mangoes at N5x. He sold each for 3x kobo and made a gain of N8. Find the value of x
Bayanin Amsa
Tambaya 38 Rahoto
Given that the root of an the equation \(2x^2 + (k+2)x+k=0\) is 2, find the value of k
Bayanin Amsa
If 2 is a root of the equation \(2x^2 + (k+2)x+k=0\), then we can substitute x = 2 into the equation and get: \[2(2)^2 + (k+2)(2) + k = 0\] Simplifying this gives: \[8 + 2k + 4 + k = 0\] \[3k = -12\] \[k = -4\] Therefore, the value of k is -4.
Tambaya 39 Rahoto
In the diagram, KL//MN, ?LKP = 30o and ?NMP = 45o. Find the size of the reflex ?KPM.
Bayanin Amsa
Tambaya 40 Rahoto
From a point P, R is 5km due west and 12km due south. Find the distance between P and R
Bayanin Amsa
To find the distance between points P and R, we can use the Pythagorean theorem which states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
In this case, we can consider P, R, and the point where the southward line from P intersects with the westward line from R, forming a right-angled triangle as shown in the diagram below:
R
*
|
|
|12 km
|
|
*------P
5 km
Let's call the point where the two lines intersect Q. Then we can see that the distance between P and Q is 5km, and the distance between Q and R is 12km. Therefore, we can use the Pythagorean theorem to find the distance between P and R as follows:
distance between P and R = sqrt(5^2 + 12^2)
= sqrt(169)
= 13 km
Therefore, the correct answer is 13km.
Tambaya 41 Rahoto
Solve the equation \(2^7 = 8^{5-x}\)
Bayanin Amsa
We know that $8 = 2^3$, so $8^{5-x} = (2^3)^{5-x} = 2^{3(5-x)} = 2^{15-3x}$. Therefore, the equation becomes: $$2^7 = 2^{15-3x}$$ Since the bases are equal, we can equate the exponents: $$7 = 15-3x$$ Solving for $x$ gives: $$x = \frac{15-7}{3} = \frac{8}{3}$$ Therefore, the solution is $\frac{8}{3}$.
Tambaya 42 Rahoto
An arc of a circle of radius 14cm subtends angle 300o at the center. Find the perimeter of the sector formed by the arc
Bayanin Amsa
To find the perimeter of the sector, we need to find the length of the arc and the length of the two radii that form the sector. The circumference of a circle with radius 14cm is given by: C = 2πr = 2π(14) = 28π Therefore, the length of the arc subtending an angle of 300 degrees is: L = (300/360) × 28π = 7π/3 × 28 ≈ 73.33cm The length of the two radii that form the sector is: 2r = 2 × 14 = 28cm Therefore, the perimeter of the sector is: P = L + 2r ≈ 73.33 + 28 = 101.33cm Hence, the correct option is (c) 101.33cm.
Tambaya 43 Rahoto
The angle of elevation of the top of a cliff 15 meters high from a landmark is 60o. How far is the landmark from the foot of the cliff? Leave your answer in surd form
Bayanin Amsa
Tambaya 44 Rahoto
Make f the subject of the relation \(v = u + ft\)
Bayanin Amsa
To make `f` the subject of the relation `v = u + ft`, we need to isolate `f` on one side of the equation. Starting with `v = u + ft`, we can begin by subtracting `u` from both sides to get: `v - u = ft` Then, dividing both sides by `t`, we get: `f = (v - u)/t` Therefore, the correct answer is: - `(v-u)/t`
Tambaya 45 Rahoto
Simplify: \(\frac{2}{3xy} - \frac{3}{4yz}\)
Bayanin Amsa
To simplify the expression \(\frac{2}{3xy} - \frac{3}{4yz}\), we need to find a common denominator and combine the two fractions.
The common denominator of the two fractions is 12xyz, which is the lowest multiple of 3xy and 4yz. We can convert each fraction to an equivalent fraction with the common denominator of 12xyz as follows:
2 3 4
-- - -- = --
3xy 4yz 12xyz
To find the numerators of the equivalent fractions, we multiply each numerator by the missing factor in the denominator of the other fraction. For example, we multiply the numerator of the first fraction by 4z to get a denominator of 12xyz, and we multiply the numerator of the second fraction by 3x to get a denominator of 12xyz:
2(4z) 3(3x) 8z - 9x
------ - ------ = --------
3xy(4z) 4yz(3x) 12xyz
Now we can simplify the expression by combining the two fractions:
8z - 9x
--------
12xyz
Therefore, the simplified expression is \(\frac{8z-9x}{12xyz}\), which corresponds to.
Therefore, the correct answer is: \(\frac{8z-9x}{12xyz}\).
Tambaya 46 Rahoto
which of the following statements describes the locus of a point R which moves in a plane such that its equidistant from two intersecting lines?
Bayanin Amsa
The locus of a point R which moves in a plane such that it is equidistant from two intersecting lines is a bisector of the angle formed by the lines. This is because a point equidistant from two intersecting lines is located at the same distance from each line, and the only points that satisfy this condition lie on the bisector of the angle formed by the lines. Therefore, the correct option is: the bisector of the angle formed by the lines.
Tambaya 47 Rahoto
In the diagram, PR is a diameter, ?PRQ = (3x-8)o and ?RPQ = (2y-7)o. s x in terms of y
Bayanin Amsa
Tambaya 48 Rahoto
The sketch shows a plot of land .
(a) Using a scale of 1 cm to 10m, draw an accurate diagram of the plot ;
(b) Construct : (i) The locus \(l_{1}\) of points equidistant from AC and BC ; (ii) the locus \(l_{2}\) of points 60m from A.
(c) A tree T inside the plot is on both \(l_{1}\) and \(l_{2}\). Locate T and find |TC| in metres.
(d) A flagpole, P is to be placed such that it it is nearer AC than BC and more than 60m from A. Shade the regions where P can be located.
Tambaya 49 Rahoto
(a) Copy and complete the table.
\(y = x^{2} - 2x - 2\) for \(-4 \leq x \leq 4\)
| x | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
| y | 22 | -2 | 1 | 6 |
(b) Using a scale of 2 cm to 1 unit on the x- axis and 2 cm to 5 units on the y- axis, draw the graph of \(y = x^{2} - 2x - 2\).
(c) Use your graph to find : (i) the roots of the equation \(x^{2} - 2x - 2 = 0\) ; (ii) the values of x for which \(x^{2} - 2x - 4\frac{1}{2} = 0\) ; (iii) the equation of the line of symmetry of the curve.
Tambaya 50 Rahoto
(a) Simplify : \(\frac{5}{8} of 2\frac{1}{2} - \frac{3}{4} \div \frac{3}{5}\).
(b) A cone and a right pyramid have equal heights and volumes. If the area of the base of the pyramid is \(154 cm^{2}\), find the base radius of the cone. [Take \(\pi = \frac{22}{7}\)].
None
Bayanin Amsa
None
Tambaya 51 Rahoto
(a) A regular polygon of n sides is such that each interior angle is 120° greater than the exterior angle. Find :
(i) the value of n ; (ii) the sum of all the interior angles.
(b) A boy walks 6km from a point P to a point Q on a bearing of 065°. He then walks to a point R, a distance of 13km, on a bearing of 146°.
(i) Sketch the diagram of his movement. (ii) Calculate, correct to the nearest kilometre, the distance PR.
Tambaya 52 Rahoto
(a) Evaluate without using the mathematical table or calculator, \(\log_{10} \sqrt{30} - \log_{10} \sqrt{6} + \log_{10} \sqrt{2}\).
(b) 
\(U = {1, 2, 3, ..., 10} ; A = {1, 2, 3, 4, 5} ; B = {2, 3, 5}\) and \(C = {6, 8, 10}\). (i) Given that the Venn diagram represents the sets above, copy and fill in the elements.
(ii) Find \(A \cap C\) ; (iii) Find \(A \cap B'\).
Bayanin Amsa
None
Tambaya 53 Rahoto
(a) Find the smallest integer that satisfies the inequality \(x + 8 < 4x - 15\).
(b) A sales girl is paid a monthly salary of N2,500 in addition to a commission of 5 kobo in the naira on all sales made by her during the month. If her sales for a month amounts to N200,000.00, calculate her income for that month.
(c) 
The diagram shows a window consisting of a rectangular and semi- circular parts. The radius of the semi- circular part is 35 cm and the height of the rectangular part is 50 cm. Find the area of the window. [Take \(\pi = \frac{22}{7}\)].
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Bayanin Amsa
None
Tambaya 54 Rahoto
(a) Two fair die are thrown once. Find the probabitlity of getting : (i) the same digit ; (ii) a total score greater than 5.
(b) Given that \(x = \cos 30°\) and \(y = \sin 30°\), evaluate without using a mathematical table or calculator : \(\frac{x^{2} + y^{2}}{y^{2} - x^{2}}\).
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Bayanin Amsa
None
Tambaya 55 Rahoto
(a) Evaluate : \(2 \div (\frac{64}{125})^{-\frac{2}{3}}\)
(b) The lines \(y = 3x + 5\) and \(y = - 4x - 1\) intersect at a point k. Find the coordinates of k.
Bayanin Amsa
None
Tambaya 56 Rahoto
(a) A plane flies due East from A(lat. 53°N, long. 25°E) to a point B(lat. 53°N, long. 85°E) at an average speed of 400 km/h. The plane then flies South from B to a point C 2000km away. Calculate, correct to the nearest whole number :
(a) the distance between A and B.
(b) the time the plane takes to reach point B ;
(c) the latitude of C.
[Take radius of the earth = 6400km; \(\pi = \frac{22}{7}\)].
Bayanin Amsa
None
Tambaya 57 Rahoto
(a) Without using mathematical table or calculator, evaluate : \(\sqrt{\frac{0.18 \times 12.5}{0.05 \times 0.2}}\).
(b) Simplify : \(\frac{8 - 4\sqrt{18}}{\sqrt{50}}\).
(c) x, y and z are related such that x varies directly as the cube of y and inversely as the square of z. If x = 108 when y = 3 and z = 4, find z when x = 4000 and y = 10.
Bayanin Amsa
None
Tambaya 58 Rahoto
The table shows the age distributions of the members of a club.
| Age (years) | 10-14 | 15-19 | 20-24 | 25-29 | 30-34 | 35-39 |
| Frequency | 7 | 18 | 25 | 17 | 9 | 4 |
(a) Calculate, correct to one decimal place, the mean age.
(b) (i) Draw a histogram to illustrate the information.
(ii) Use the histogram to estimate the modal age .
(c) If a member is selected at random, what is the probability that he/she is in the modal class?
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Bayanin Amsa
None
Tambaya 59 Rahoto
(a) 
In the diagram, A, B, C and D are points on the circumference of a circle. XY is a tangent at A. Find : (i) < CAX ; (ii) < ABY.
(b) If (m + 1) and (m - 3) are factors of \(m^{2} - km + c\), find the values of k and c.
Bayanin Amsa
None
Tambaya 60 Rahoto
(a) 
In the diagram, \(\Delta\) ABD is right-angled at B. |AB| = 3 cm, |AD| = 5 cm, \(\stackrel\frown{ACB}\) = 61° and \(\stackrel\frown{DAC}\) = x°. Calculate, correct to one decimal place, the value of x.
(b) 
In the diagram, OABCD is a pyramid with a square base of side 2cm and a slant height of 4 cm. Calculate, correct to three significant figures : (i) the vertical height of the pyramid ; (ii) the volume of the pyramid.
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Bayanin Amsa
None
